What is the extreme f(x,y)=x^3-12x+y^3+3y^2-9y ?

The extreme f(x,y)=x^3-12x+y^3+3y^2-9y is Saddle(-2,1),Maximum(-2,-3),Minimum(2,1),Saddle(2,-3)

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extreme f(x,y)=x^3-12x+y^3+3y^2-9y

extreme points

f (x, y) = x^{3} - 12 x + y^{3} + 3 y^{2} - 9 y

Saddle (- 2, 1), Maximum (- 2, - 3), Minimum (2, 1), Saddle (2, - 3)

Solution steps

Find the critical points:

(- 2, 1), (- 2, - 3), (2, 1), (2, - 3)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y + 6

D (x, y) = 6 x (6 y + 6)

Check the sign of at each critical point

Check critical point

(- 2, 1) :

Saddle

Check critical point

(- 2, - 3) :

Maximum

Check critical point

(2, 1) :

Minimum

Check critical point

(2, - 3) :

Saddle

Saddle (- 2, 1), Maximum (- 2, - 3), Minimum (2, 1), Saddle (2, - 3)

f (x, y) = 4 x^{2} - 2 y^{2} + 2 x y

f (x) = a e^{2 x} - e^{3 x}

f (x) = x^{4} + 8 x^{3} + 18 x^{2} + 4

f (x) = x^{4} + 8 x^{3} + 18 x^{2} - 8

f (x, y) = 4 x y - x^{3} - 2 y^{2}

What is the extreme f(x,y)=x^3-12x+y^3+3y^2-9y ?
The extreme f(x,y)=x^3-12x+y^3+3y^2-9y is Saddle(-2,1),Maximum(-2,-3),Minimum(2,1),Saddle(2,-3)